Question

can someone help me with a trig equation?

Answer 1

im mirin your pic

Answer 2

\[\cos(2\theta-\pi/2)=-1\]

Answer 3

what about it.

Answer 4

its a trig question, can you help me solve it

Answer 5

in what interval

Answer 6

im not sure what you mean

Answer 7

trig equations have an infinite amount of solutions.

are you asking for a general solution or a set of solutions within a specific interval?

Answer 8

well i kind of need to know how to isolate the cos first the interval is 0<theta<2

Answer 9

is it (2theta-pi) / 2
or 2theta - (pi/2) ?

Answer 10

the second one

Answer 11

use cos(a-b) = cosacosb +sinasinb

Answer 12

where did you get the 0?

Answer 13

cos (2theta - (pi/2) = -1

let 2theta - (pi/2) = l

cos l = -1

l = pi

2theta - (pi/2) = pi

theta = [pi + (pi/2)] / 2

Therefore, theta = [pi + (pi/2)] / 2 + kpi where k is an integer.

to solve for the interval [0,2pi], add or subtract a period accordingly.

Answer 14

cos(pi/2) = 0 so multiple by a number = 0

Answer 15

thank you, can you explain what you did with the 2theta-pi/2=1?

Answer 16

why did you substitute pi with 1?

Answer 17

where did i substitute pi with 1?

Answer 18

oops, i got confused sorry. what did you do with the 2theta-pi/2=1

Answer 19

I think the easy way is to use cos(a-b) = cosacosb+sinasinb

Answer 20

i made it cos l = -1, solved for l and back substituted 2theta - (pi/2) to solve for theta.

Answer 21

im sorry, what do you mean back substituted? I just barely learned the unit circle, so im trying to understand all this lol, sorry

Answer 22

since l = 2theta - (pi/2)

and i solved for cos l = -1

to be l = pi

thits means 2theta - (pi/2) = pi also

to solve for theta, i simply isolated theta.

Answer 23

thank you